The document discusses column stability and Euler's formula for pin-ended columns. It provides: - An introduction to column stability and buckling under compressive loads. - A model of a column as two rods with a torsional spring to explain buckling behavior. - Derivation of Euler's formula which gives the critical buckling load of a pin-ended column as a function of its properties. - Discussion of how column geometry, specifically its slenderness ratio, affects buckling behavior.

1. Column
Introduction , Stability of structures, Euler’s
formula

2.  Introduction
 Stability of structures
 Euler's formula for pin-ended columns
Columns

3. • In the design of columns, cross-sectional area is
selected such that
- allowable stress is not exceeded
all
A
P

 

- deformation falls within specifications
spec
AE
PL

 

• After these design calculations, may discover
that the column is unstable under loading and
that it suddenly becomes sharply curved or
buckles.
Stability of Structures

4. • Long slender members subjected to axial
compressive force are called columns.
• The lateral deflection that occurs is
called buckling.
• The maximum axial load a column can
support when it is on the verge of
buckling is called the critical load, Pcr.
Stability of Structures

5. • Consider model with two rods and torsional
spring. After a small perturbation,
 
moment
ing
destabiliz
2
sin
2
moment
restoring
2









L
P
L
P
K
• Column is stable (tends to return to
aligned orientation) if
 
L
K
P
P
K
L
P
cr
4
2
2




 

Stability of Structures

6. • Assume that a load P is applied. After a
perturbation, the system settles to a new
equilibrium configuration at a finite
deflection angle.
 




sin
4
2
sin
2



cr
P
P
K
PL
K
L
P
• Noting that sinθ < θ, the assumed
configuration is only possible if P > Pcr.
Stability of Structures

7. • Summing moments, M = Py, equation of
bending becomes
• General solution is
• Since y = 0 at x = 0, then B = 0.
Since y = 0 at x = L, then
0
2
2







 y
EI
P
dx
y
d

















 x
EI
P
cos
B
x
EI
P
sin
A
y
0









L
EI
P
sin
A
Euler’s Formula for Pin-Ended Beams

8. • Disregarding trivial solution for A = 0, we get
• Which is satisfied if
or
,...
3
,
2
,
1
2
2
2

 n
L
EI
n
P


n
L
EI
P

0
sin 






L
EI
P
Euler’s Formula for Pin-Ended Beams

9. • Smallest value of P is obtained for n = 1, so critical load for column
is
• This load is also referred to
as the Euler load. The
corresponding buckled
shape is defined by
• A represents maximum
deflection, ymax, which occurs
at midpoint of the column.
2
2
L
EI
Pcr


L
x
sin
A
y


Euler’s Formula for Pin-Ended Beams

10. • A column will buckle about the principal axis of the x-
section having the least moment of inertia (weakest axis).
• For example, the meter stick shown will
buckle about the a-a axis and not
the b-b axis.
• Thus, circular tubes made excellent
columns, and square tube or those
shapes having Ix ≈ Iy are selected
for columns.
Euler’s Formula for Pin-Ended Beams

11. • Buckling equation for a pin-supported long slender column,
Pcr = critical or maximum axial load on column just before it
begins to buckle.
E = modulus of elasticity of material
I = Least modulus of inertia for column’s x-sectional area.
L = unsupported length of pinned-end columns.
2
2
L
EI
Pcr


Euler’s Formula for Pin-Ended Beams

12. • The value of stress corresponding to the critical load,
A
L
EI
A
P
L
EI
P
cr
cr
cr
cr
2
2
2
2







Slenderness Ratio

13. Expressing I = Ar2 where A is x-sectional area of column and
r is the radius of gyration of x-sectional area.
cr = critical stress, an average stress in column just before
the column buckles.
E = modulus of elasticity of material
L = unsupported length of pinned-end columns.
r = smallest radius of gyration of column
 2
2
r
L
E
cr

 
Slenderness Ratio

14. • The geometric ratio L/r in the equation is known as the
slenderness ratio.
• It is a measure of the column’s flexibility and will be used to
classify columns as long, intermediate or short.
Slenderness Ratio

15. IMPORTANT
• Columns are long slender members that are subjected to
axial loads.
• Critical load is the maximum axial load that a column can
support when it is on the verge of buckling.
• This loading represents a case of neutral equilibrium.
Ideal Column with Pin Supports

16. IMPORTANT
• An ideal column is initially perfectly straight, made of
homogeneous material, and the load is applied through the
centroid of the x-section.
• A pin-connected column will buckle about the principal axis
of the x-section having the least moment of intertia.
• The slenderness ratio L/r, where r is the smallest radius of
gyration of x-section. Buckling will occur about the axis
where this ratio gives the greatest value.
• Preceding analysis is limited to centric loadings.
Ideal Column with Pin Supports

17. Columns Having Various Types of Supports

18. Determine the critical load of an aluminium tube
that is 1.5 m long and has 16 mm outer diameter
and 1.25 mm wall thickness. Use E=70 GPa.
Example

19. Solution:
Example

20. The A-36 steel W20046 member shown is to be used as a
pin-connected column. Determine the largest axial load it
can support before it either begins to
buckle or the steel yields.
Example

21. From calculation, column’s x-sectional area and
moments of inertia are A = 5890 mm2,
Ix = 45.5106 mm4,and Iy = 15.3106 mm4.
By inspection, buckling will occur about the y-y axis.
Applying buckling equation, we have
 
   
  
 
kN
6
.
1887
m
4
mm
1000
/
m
1
mm
10
3
.
15
kN/m
10
200
2
4
4
4
2
6
2
2
2





L
EI
Pcr
Example

22. When fully loaded, average compressive stress in column is
Since this stress exceeds yield stress (250 N/mm2), the load P
is determined from simple compression:
kN
5
.
1472
mm
5890
N/mm
250 2
2


P
P
 
2
2
N/mm
5
.
320
mm
5890
N/kN
1000
kN
6
.
1887



A
Pcr
cr

Example

23. Page 619 No 10.1
Example

24. Page 619 No 10.2
Example

25. Page 620 No 10.10
Example

26. Page 621 No 10.13
Example

27. Example
Page 621 No 10.12

28. Example
Page 621 No 10.12

29. Example

30. Example
Solution
Page 624 No 10.27

31. Solution

32. Exercise 1
Determine
(a) The crtical load for the steel strut
(b) The dimension d for which the aluminium strut will have same critical load
(c) Express the weight of the aluminium alloy strut as a percent of the weight of
the steel strut.
Refer to Figure No
10.12 Page 621
Steel
E = 200 GPa
Ρ = 7860 kg/m3

33. Exercise 2
A compression member of 1.5 m effective length consists of a solid 30 mm
diameter brass rod. In order to reduce the weight of the member by 25%, the
solid rod is replaced by a hollow rod of the cross section shown. Determine
(a) the percent reduction in the critical load
(b) The value of the critical load for the hollow rod. Use E = 70 GPa
Refer to Figure No
10.13 Page 621
15
mm
30 mm
30 mm

34. Exercise 3

35. Solution

36. Exercise 4

37. Solution